Post on 13-Nov-2014
CSE 260
BRAC University, Fall’06
Drawing Logic Circuit
When a Boolean expression is provided, we can easily draw the logic circuit.
Examples: (i) F1 = xz' (note the use of a 2-input AND gate)
x
z
F1
z'
Drawing Logic Circuit
(ii) F2 = x + y'z (can assume that variables and their complements are available)
(iii) F3 = xy' + x'z
x
y'z
F2
y'z
x'z
F3
x'z
xy'xy'
Solve it yourself
Q1. Draw a logic circuit for BD + BE + D’F
Q2. Draw a logic circuit for
A’BC + B’CD + BC’D + ABD’
Analysing Logic Circuit When a logic circuit is provided, we can
analyse the circuit to obtain the logic expression.
Example: What is the Boolean expression of F4?
A'B'A'B'+C (A'B'+C)'
A'
B'
CF4
F4 = (A'B'+C)' = (A+B).C'
Solve it yourself
What is Boolean expression of F5?
z
F5
x
y
Complement of Functions
Given a function, F, the complement of this function, F', is obtained by interchanging 1 with 0 in the function’s output values.
x y z F1 F1'0 0 0 0 10 0 1 0 10 1 0 0 10 1 1 0 11 0 0 0 11 0 1 0 11 1 0 1 01 1 1 0 1
Example: F1 = xyz'
Complement: F1' = (xyz')' = x' + y' + (z')' DeMorgan = x' + y' + z Involution
Complement of Functions
More general DeMorgan’s theorems useful for obtaining complement functions:
(A + B + C + ... + Z)' = A' . B' . C' … . Z'
(A . B . C ... . Z)' = A' + B' + C' + … + Z'
Standard Forms Two Standard Forms:
Sum-of-Products and Product-of-Sums
Literals: a variable on its own or in its complemented form. Examples: x, x' , y, y'
Product Term: a single literal or a logical product (AND) of several literals.
Examples: x, xyz', A'B, AB
Standard Forms
Sum Term: a single literal or a logical sum (OR) of several literals.Examples: x, x+y+z', A'+B, A+B
Sum-of-Products (SOP) Expression: a product term or a logical sum (OR) of several product terms.
Examples: x, x+yz', xy'+x'yz, AB+A'B' Product-of-Sums (POS) Expression: a sum term or a logical product
(AND) of several sum terms.Exampes: x, x(y+z'), (x+y')(x'+y+z), (A+B)(A'+B')
Every boolean expression can either be expressed as sum-of-products or product-of-sums expression.
Examples:SOP: xy + xy + xyzPOS: (x + y)(x + y)(x + z)
Minterm
Consider two binary variables x, y.
Each variable may appear as itself or in complemented form as literals (i.e. x, x' & y, y' )
For two variables, there are four possible combinations with the AND operator, namely:
x'y', x'y, xy', xy
These product terms are called the minterms.
A minterm of n variables is the product of n literals from the different variables.
In general, n variables can give 2n minterms.
Maxterm
In a similar fashion, a maxterm of n variables is the sum of n literals from the different variables.
Examples: x'+y', x'+y, x+y',x+y
In general, n variables can give 2n maxterms.
Minterm & Maxterm
The minterms and maxterms of 2 variables are denoted by m0 to m3 and M0 to M3 respectively:
Minterms Maxtermsx y term notation term notation0 0 x'y' m0 x+y M00 1 x'y m1 x+y' M11 0 xy' m2 x'+y M21 1 xy m3 x'+y' M3
Each minterm is the complement of the corresponding maxterm: Example: m2 = xy'
m2' = (xy')' = x' + (y')' = x'+y = M2
Canonical Form: Sum of Minterms
Obtain Sum-of-Minterms by gathering/summing the minterms of the function (where result is a 1)F1 = xyz' = (m6)F2 = x'y'z+xy'z'+xy'z+xyz'+xyz = (m1,m4,m5,m6,m7)F3 = x'y'z+x'yz+xy'z'
+xy'z = (m1,m3,m4,m5)
x y z F1 F2 F30 0 0 0 0 00 0 1 0 1 10 1 0 0 0 00 1 1 0 0 11 0 0 0 1 11 0 1 0 1 11 1 0 1 1 01 1 1 0 1 0
Canonical Form: Product of Maxterms
Maxterms are sum terms.
For Boolean functions, the maxterms of a function are the terms for which the result is 0.
Boolean functions can be expressed as Products-of-Maxterms.
Canonical Form: Product of Maxterms
E.g.: F2 = (M0,M2,M3) = (x+y+z)(x+y'+z)(x+y'+z')
F3 = (M0,M2,M6,M7)
= (x+y+z)(x+y'+z)(x'+y'+z)(x'+y'+z') x y z F1 F2 F30 0 0 0 0 00 0 1 0 1 10 1 0 0 0 00 1 1 0 0 11 0 0 0 1 11 0 1 0 1 11 1 0 1 1 01 1 1 0 1 0
Canonical Form: Product of Maxterms
Why is this so? Take F2 as an example.F2 = (m1,m4,m5,m6,m7)
The complement function of F2 is:F2' = (m0,m2,m3) = m0 + m2 + m3
Therefore:F2 = (m0 + m2 + m3 )' = m0' . m2' . m3' DeMorgan = M0 . M2 . M3 mx' = Mx = (M0,M2,M3)
Every Boolean function can be expressed as either Sum-of-Minterms or Product-of-Maxterms.
x y z F2 F2'0 0 0 0 10 0 1 1 00 1 0 0 10 1 1 0 11 0 0 1 01 0 1 1 01 1 0 1 01 1 1 1 0
Conversion of Canonical Forms
Sum-of-Minterms Product-of-Maxterms Rewrite minterm shorthand using maxterm shorthand. Replace minterm indices with indices not already used.
Eg: F1(A,B,C) = m(3,4,5,6,7) = M(0,1,2)
Product-of-Maxterms Sum-of-Minterms Rewrite maxterm shorthand using minterm shorthand. Replace maxterm indices with indices not already used.
Eg: F2(A,B,C) = M(0,3,5,6) = m(1,2,4,7)
Solving it yourself1. A Boolean function of 5 variables can have up to ___
minterms.
2. Given a Boolean function F (A,B,C) = m(0, 5, 7).Which of the following is correct?a) F' (A,B,C) = m(0,5,7) b) F (A,B,C) = M(1,2,3,4,6)c) F (A,B,C) = M(0,5,7) d) F' (A,B,C) = m(1,2,3)e) None above
3. Given a Boolean function F (x,y,z) = y'.(x + z') + x'.z.Which of the following is correct?a) F (x,y,z) = m(0,1) b) F (x,y,z) = m(0,1,4,5)c) F (x,y,z) = m(0,1,2,3,4) d) F (x,y,z) = m(0,1,3,4,5)e) F (x,y,z) = m(1,2,3,4,5)
Solving it yourself
4. Given a Boolean function with 6 variables a, b, c, d, e, f.What is maxterm M60?a) a'.b'.c'.d'.e.f b) a'+b'+c'+d' c) a'+b'+c'+d'+e+fd) a.b.c.d.e'.f' e) a+b+c+d+e'+f'
5. Given a Boolean function F (a,b,c) = S m(0, 2, 5, 6, 7).If a=0, b=c=1, what is the value of F?
a) 0 b) 1 c) b.c d) a' e) Unknown
6. Which of the following is NOT a minterm of the Boolean function: F (w,x,y,z) = w.x.z' + x.y'.z + x.z
a) w.x.y.z' b) w'.x.y'.z c) w.x.y.z
d) w.x.y'.z' e) w.x'.y.z
Function Simplification Why simplify?
Simpler expression uses less logic gates. Thus: cheaper, less power, faster (sometimes).
Algebraic Simplification Example:(x+y)(x+y')(x'+z) (6 literals)
= (xx+xy'+xy+yy')(x'+z) (assoc.)
= (x+x(y'+y)+0)(x'+z) (idemp.,assoc., compl.)
= (x+x(1)+0)(x'+z) (complement)
= (x+x+0)(x'+z) (identity 1)
= (x)(x'+z) (idemp, identity 0)
= (xx'+xz) (assoc.)
= (0+xz) (complement)
= xz (identity 0)
Number of literals reduced from 6 to 2.
Algebraic Simplification Find minimal SOP expression of
f(a,b,c,d) = abc + abd + a'bc' + cd + bd'abc + abd + a'bc' + cd + bd'= abc + ab + a'bc' + cd + bd' (absorption)= abc + ab + bc' + cd + bd' (absorption)= ab + bc' + cd + bd' (absorption)= ab + cd + b(c' + d') (distributivity)= ab + cd + b(cd)' (DeMorgan)= ab + cd + b (absorption)= b + cd (absorption)
Number of literals reduced form 13 to 3.
Assignment
2-13, 2-14, 2-16, 2-18